Question

If \( y = e^{\tan^{-1}x} \), prove that \( (1 + x^2)\frac{d^2y}{dx^2} + (2x - 1)\frac{dy}{dx} = 0 \).

Hint:

When asked to prove a differential equation, after finding the first derivative, it's often a good strategy to rearrange the equation to eliminate fractions or complex terms (like substituting 'y' back in). Differentiating this simpler, rearranged form can make finding the second derivative much easier and lead directly to the required result.

Answer 1

Step 1: Understanding the Concept:
This problem requires showing that a given function satisfies a second-order linear differential equation. The process involves finding the first and second derivatives of the function and substituting them into the equation to verify the identity.
Step 2: Key Formula or Approach:
We will use the chain rule and product rule for differentiation.
- Chain Rule: \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \)
- Product Rule: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \)
- Derivative of \( e^u \): \( \frac{d}{dx}e^u = e^u \frac{du}{dx} \)
- Derivative of \( \tan^{-1}x \): \( \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2} \)
Step 3: Detailed Explanation or Calculation:
Given the function \( y = e^{\tan^{-1}x} \).
First Derivative:
Differentiate y with respect to x using the chain rule:
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{\tan^{-1}x}) = e^{\tan^{-1}x} \cdot \frac{d}{dx}(\tan^{-1}x) \] \[ \frac{dy}{dx} = e^{\tan^{-1}x} \cdot \frac{1}{1+x^2} \] Since \( y = e^{\tan^{-1}x} \), we can substitute y back into the equation:
\[ \frac{dy}{dx} = \frac{y}{1+x^2} \] To avoid fractions when differentiating again, rearrange this equation:
\[ (1+x^2)\frac{dy}{dx} = y \] Second Derivative:
Now, differentiate both sides of the rearranged equation with respect to x, using the product rule on the left side:
\[ \frac{d}{dx}\left((1+x^2)\frac{dy}{dx}\right) = \frac{d}{dx}(y) \] \[ \left(\frac{d}{dx}(1+x^2)\right)\frac{dy}{dx} + (1+x^2)\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{dy}{dx} \] \[ (2x)\frac{dy}{dx} + (1+x^2)\frac{d^2y}{dx^2} = \frac{dy}{dx} \] Now, rearrange the terms to match the required equation:
\[ (1+x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} - \frac{dy}{dx} = 0 \] Factor out \( \frac{dy}{dx} \) from the last two terms:
\[ (1+x^2)\frac{d^2y}{dx^2} + (2x - 1)\frac{dy}{dx} = 0 \] Step 4: Final Answer:
We have successfully derived the given differential equation starting from the function \( y = e^{\tan^{-1}x} \). Hence, the relation is proved.